Find the points on the curve $y = 4 x^{3} - 3 x + 5$ at which the equation of the tangent is parallel to the x-axis.

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Solution

Equation of the tangent is parallel to the x-axis $\to$ Slope of tangent $m = 0$ .
Differentiate $y$ w.r.t $x$
$\frac{d y}{d x} = m = 12 x^{2} - 3$
$m = 0 = 12 x^{2} - 3 ⇛ x^{2} = \frac{1}{4}$
$x = - \frac{1}{2}, \frac{1}{2}$
Corresponding $y$ will be,
$x = - \frac{1}{2} \Rightarrow y = 4 (- \frac{1}{2})^{3} - 3 (- \frac{1}{2}) + 5 = 6$
$(- \frac{1}{2}, 6)$
$x = \frac{1}{2} \Rightarrow y = 4 (\frac{1}{2})^{3} - 3 (\frac{1}{2}) + 5 = 4$
$(\frac{1}{2}, 4)$

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